$\ce{C2H5+ vs \ C3H7+}$ There are two conflicting trends here.

Inductive effect of ethyl will be higher than that of methyl so carbocation attached to ethyl (3 carbons in total) should be more stable.

On the other hand 3 hyperconjugation structures can be drawn for the carbocation attached to methyl (2 carbons in total) and only 2 hyperconjugation structure for the other.

Which effect wins? I searched a lot of books but didn't find anything.


2 Answers 2


The ethyl carbocation ($\ce{C2H5+}$) has a different structure than what you might have anticipated. It is a non-classical ion with a bridging hydrogen. For a description of just what a non-classical carbocation is see this answer. The hallmark of a non-classical ion is 3 atoms with 2 electrons spread over them. This is called a 3-center 2-electron bond (hypercoordinate bonding). In the case at hand the 3-atom 2-electron bond involves the 2 carbon atoms and the bridged hydrogen atom.

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So there is no "methyl" group present in the ethyl carbocation that can inductively release electrons. Additionally, the hyperconjugated resonance structure you mentioned has actually been replaced by a hydrogen that is partially bonded to the other carbon atom.

The n-propyl carbocation would likely have a similar non-classical structure - if it existed at all. This carbocation has not been experimentally observed, it rearranges too rapidly to the much more stable 2-propyl carbocation.

So while we can't answer your question about whether inductive or hyperconjugative effects are more important in this series of compounds, we can assess which cation is more stable.

A variety of experiments suggest that the heat of formation of the non-classical ethyl cation is around 215 kcal/mol (reference, see pp. 68-72). This same reference points out that the heat of formation of the classical 1-propyl cation lies approximately 20 kcal/mol above that of the 2-propyl cation. The heat of formation of the 2-propyl cation is around 193 kcal/mol. Adding 20 kcal/mol to that value would place the classical 1-propyl cation around 213 kcal/mol. But, assuming the 1-propyl cation would also be a non-classical ion it would be lower in energy than the classical version, its heat of formation would therefore be less than 213 kcal/mol.

This analysis suggests that the 1-propyl cation would likely have a lower heat of formation than the ethyl cation. In other words, the 1-propyl cation would likely be more stable than the ethyl cation.

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    $\begingroup$ But what is the reason behind the increased stability of the 1-propyl cation over the ethyl cation? With reference to an inductive argument, does a longer chain mean a "larger stock of electrons" to release to stabilise the carbocation? $\endgroup$ Commented Dec 29, 2018 at 9:20

You never defined which carbon is the carbocation in the case of the propylium carbocation. I will go through a comparison of all three possibilities: the case of the ethylium carbocation, the linear propylium carbocation with a terminal carbocation, and the linear propylium carbocation with an internal carbocation.

Thermodynamically: Ethylium < terminal propylium < internal propylium

Kinetically: Terminal propylium < ethylium < internal propylium

The difference in the kinetic and thermodynamic stability arises from the fact that the terminal propylium will readily undergo a 1,2-hydride shift to create the internal propylium ion.

As for the thermodynamic order, it gets a bit more complex. Remember that for alkanes, it is ALWAYS the case that the more substituted carbocation is more stable and in cases of ties, more substituted substituents make for more stable carbocations. As for why this is true, well it comes down to the fact that carbons are electron donating to carbocations (and other electron deficient carbons). There is a role from hyperconjugation but you are forgetting that you can have hyperconjugation from bonds other than $\ce{C-H}$. You also seem to think that more hyperconjugatory structures is better when really that is not the case. Hyperconjugation is NOT an example of resonance where the molecule is in a superposition of multiple states so there is no enthalpic benefit of more methods of hyperconjugation. There is an entropic effect from the additional degrees of freedom but this would be a very small difference and if I remember right, would actually benefit both propylium carbocations, not the ethylium carbocation.

  • $\begingroup$ Why isn't hyperconjugation an example of resonance, there is no nuclear movement in such structures? $\endgroup$
    – ron
    Commented Apr 19, 2015 at 23:59
  • $\begingroup$ I'm sorry for not making it clear but I mean the linear propylium carbocation with a terminal carbocation. $\endgroup$ Commented Apr 20, 2015 at 11:14
  • $\begingroup$ Also all books I've read state that more hyperconjugation structures imply more stability. March in his book for example , gives this as the reason why secondary carbocations are more stable than primary ones.(Along with inductive effect explanation). $\endgroup$ Commented Apr 20, 2015 at 11:17
  • $\begingroup$ See chemistry.stackexchange.com/questions/4151/… for another example. $\endgroup$ Commented Apr 20, 2015 at 11:19

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